A) \[A=0,\,B=1\]
B) \[A=1,\,B=1\]
C) \[A=-1,\,B=1\]
D) \[A=-1,\,B=0\]
Correct Answer: C
Solution :
For continuity at all \[x\in R,\] we must have \[f\left( -\frac{\pi }{2} \right)=\underset{x\to {{(-\pi /2)}^{-}}}{\mathop{\text{lim}}}\,(-2\sin x)\]\[=\underset{x\to {{(-\pi /2)}^{+}}}{\mathop{\text{lim}}}\,(A\sin x+B)\] Þ \[2=-A+B\] .....(i) and \[f\left( \frac{\pi }{2} \right)=\underset{x\to {{(\pi /2)}^{-}}}{\mathop{\lim }}\,(A\sin x+B)\]\[=\underset{x\to {{(\pi /2)}^{+}}}{\mathop{\text{lim}}}\,(\cos x)\] Þ \[0=A+B\] .....(ii) From (i) and (ii), \[A=-1\] and \[B=1\].
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